Question: $ B = \left[\begin{array}{rrr}1 & 4 & 1 \\ -2 & 2 & 3\end{array}\right]$ $ F = \left[\begin{array}{rr}1 & 5 \\ 3 & -1 \\ 0 & 0\end{array}\right]$ What is $ B F$ ?
Answer: Because $ B$ has dimensions $(2\times3)$ and $ F$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ B F = \left[\begin{array}{rrr}{1} & {4} & {1} \\ {-2} & {2} & {3}\end{array}\right] \left[\begin{array}{rr}{1} & \color{#DF0030}{5} \\ {3} & \color{#DF0030}{-1} \\ {0} & \color{#DF0030}{0}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ B$ , with the corresponding elements in column $j$ of the second matrix, $ F$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ B$ with the first element in ${\text{column }1}$ of $ F$ , then multiply the second element in ${\text{row }1}$ of $ B$ with the second element in ${\text{column }1}$ of $ F$ , and so on. Add the products together. $ \left[\begin{array}{rr}{1}\cdot{1}+{4}\cdot{3}+{1}\cdot{0} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ B$ with the corresponding elements in ${\text{column }1}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{1}\cdot{1}+{4}\cdot{3}+{1}\cdot{0} & ? \\ {-2}\cdot{1}+{2}\cdot{3}+{3}\cdot{0} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ B$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{1}\cdot{1}+{4}\cdot{3}+{1}\cdot{0} & {1}\cdot\color{#DF0030}{5}+{4}\cdot\color{#DF0030}{-1}+{1}\cdot\color{#DF0030}{0} \\ {-2}\cdot{1}+{2}\cdot{3}+{3}\cdot{0} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{1}\cdot{1}+{4}\cdot{3}+{1}\cdot{0} & {1}\cdot\color{#DF0030}{5}+{4}\cdot\color{#DF0030}{-1}+{1}\cdot\color{#DF0030}{0} \\ {-2}\cdot{1}+{2}\cdot{3}+{3}\cdot{0} & {-2}\cdot\color{#DF0030}{5}+{2}\cdot\color{#DF0030}{-1}+{3}\cdot\color{#DF0030}{0}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}13 & 1 \\ 4 & -12\end{array}\right] $